$f(x,y) = x\cos(y) + y$ What is $\dfrac{\partial f}{\partial x}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-\sin(y) + 1$ (Choice B) B $\cos(y)$ (Choice C) C $\cos(y) - x\sin(y) + 1$ (Choice D) D $-x\sin(y)$
Answer: We want to find $\dfrac{\partial f}{\partial x}$, which is the partial derivative of $f$ with respect to $x$. When we take a partial derivative with respect to $x$, we treat $y$ as if it were a constant. Let's break $f(x, y)$ down term by term. $\begin{aligned} &\dfrac{\partial}{\partial x} \left[ x\cos(y) \right] = \cos(y) \\ \\ &\dfrac{\partial}{\partial x} \left[ y \right] = 0 \end{aligned}$ Adding the terms back together, we get the partial derivative. In conclusion: $\dfrac{\partial f}{\partial x} = \cos(y) + 0 = \cos(y)$